Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 10 - Rotation - Problems: 21

Answer

$\omega=6.92\times10^{-13}rad/s$

Work Step by Step

First convert the velocity of the top of the tower to meters per second using dimensional analysis: $1.2mm/y\times\frac{1m}{1000mm}\times\frac{1y}{365days}\times\frac{1day}{24hours}\times\frac{1hour}{60min}\times\frac{1min}{60s}$ $=3.81\times10^{-11}m/s$ Determine the angular velocity of the tower by dividing the velocity of the top of the tower by the height of the tower. $\omega=\frac{v}{r}=\frac{3.81\times10^{-11}m/s}{55m}=6.92\times10^{-13}rad/s$
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